Let $T$ be a positive integer whose only digits are 0s and 1s. If $X = T \div 12$ and $X$ is an integer, what is the smallest possible value of $X$?
Answer: Since $T$ is divisible by 12, it must be divisible by both 3 and 4.  Hence, the sum of its digits is divisible by 3 and its last two digits are divisible by 4.  By inspection, we see that $T$ must end in 00 and therefore the smallest such $T$ is 11100.  Calculating, $X = \boxed{925}$.